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# Calculation of bearing loads and joint forces

This exercise addresses the following questions:

• How do you calculate joint forces?
• How do I calculate a frame with two fixed bearings?
• What are the different approaches to calculating joint forces?

The forces F1 to F4 act on a frame with two fixed bearings and a joint. Determine the bearing loads and the forces in joint C!

F1 = F4 = 10 kN, F2 = 30 kN, F3 = 60 kN, a = 1 m

## Solution

The solution can be done in different ways, here two variants are shown:

Solution variant 1 considers both areas to the left and right of the joint separately from the beginning and sets up the corresponding equations.

Solution variant 2 sets up the equilibrium conditions for the entire system as the first step, and then releases the forces of the sub-areas.

## Variant 1

The following video is in German language.

The frame will be divided into two sections to determine the forces.

Balance of forces section I
x-direction

$\tag{1} 0={F_{\mathit{Cx}}}+{F_{\mathit{Ax}}}+{F_1}$

y-direction

$\tag{2} 0={F_{\mathit{Cy}}}+{F_{\mathit{Ay}}}-{F_2}$

Moments around A (left-turning moments are positive)

$\tag{3} 0=2 {F_{\mathit{Cy}}} a-3 {F_{\mathit{Cx}}} a-{F_2} a-2 {F_1} a$

Balance of forces section II
x-direction

$\tag{4} 0=-{F_{\mathit{Cx}}}+{F_{\mathit{Bx}}}-{F_4}$

y-direction

$\tag{5} 0=-{F_{\mathit{Cy}}}+{F_{\mathit{By}}}-{F_3}$

Moments around B (left-turning moments are positive)

$\tag{6} 0=4 {F_{\mathit{Cy}}} a+3 {F_{\mathit{Cx}}} a+{F_4} a+2 {F_3} a$

Solve equation %o1 for FAx

$\tag{7} {F_{\mathit{Ax}}}=-{F_{\mathit{Cx}}}-{F_1}$

Solve equation 3 for FCx

$\tag{8} {F_{\mathit{Cx}}}=\frac{2 {F_{\mathit{Cy}}}-{F_2}-2 {F_1}}{3}$

Substitute FCx in equation 7

$\tag{9} {F_{\mathit{Ax}}}=-\frac{2 {F_{\mathit{Cy}}}-{F_2}-2 {F_1}}{3}-{F_1}$

Solve equation 6 for FCy

$\tag{10} {F_{\mathit{Cy}}}=-\frac{3 {F_{\mathit{Cx}}}+{F_4}+2 {F_3}}{4}$

Substitute FCy in equation 8

$\tag{11} {F_{\mathit{Cx}}}=\frac{-\frac{3 {F_{\mathit{Cx}}}+{F_4}+2 {F_3}}{2}-{F_2}-2 {F_1}}{3}$

... and solve for FCx

$\tag{12} {F_{\mathit{Cx}}}=-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}$

Solve equation 4 for FBx

$\tag{13} {F_{\mathit{Bx}}}={F_{\mathit{Cx}}}+{F_4}$

... and substitution of FCx yields FBx

$\tag{14} {F_{\mathit{Bx}}}={F_4}-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}$

Substitution of FCx in equation 10

$\tag{15} {F_{\mathit{Cy}}}=-\frac{-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_4}+2 {F_3}}{4}$

Solve equation 5 for FBy

$\tag{16} {F_{\mathit{By}}}={F_{\mathit{Cy}}}+{F_3}$

... and substitution of FCy

$\tag{17} {F_{\mathit{By}}}=\frac{-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_2}+2 {F_1}}{2}+{F_3}$

F_Cy in equation 9 gives

$\tag{18} {F_{\mathit{Ax}}}=\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}-{F_1}$

Solve equation 2 for FAy

$\tag{19} {F_{\mathit{Ay}}}={F_2}-{F_{\mathit{Cy}}}$

... and substitution of FCy yields the last missing parameter

$\tag{20} {F_{\mathit{Ay}}}={F_2}-\frac{-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_2}+2 {F_1}}{2}$

Summary of the results

$\tag{21} {F_{\mathit{Ax}}}=\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}-{F_1}$

$\tag{22} {F_{\mathit{Ay}}}={F_2}-\frac{-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_2}+2 {F_1}}{2}$

$\tag{23} {F_{\mathit{Bx}}}={F_4}-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}$

$\tag{24} {F_{\mathit{By}}}=\frac{-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_2}+2 {F_1}}{2}+{F_3}$

$\tag{25} {F_{\mathit{Cx}}}=-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}$

$\tag{26} {F_{\mathit{Cy}}}=\frac{-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_2}+2 {F_1}}{2}$

Inserting the numerical values ​​delivers

$\tag{27} {F_{\mathit{Ax}}}=15.56 \mathit{kN}$

$\tag{28} {F_{\mathit{Ay}}}=43.33 \mathit{kN}$

$\tag{29} {F_{\mathit{Bx}}}=-15.56 \mathit{kN}$

$\tag{30} {F_{\mathit{By}}}=46.67 \mathit{kN}$

$\tag{31} {F_{\mathit{Cx}}}=-25.56 \mathit{kN}$

$\tag{32} {F_{\mathit{Cy}}}=-13.33 \mathit{kN}$

## Variant 2

This solution is calculated here exclusively with the variables, since the numerical values ​​have already been calculated above.

Establishing the balance of forces and moments for the entire system

$\tag{1} \sum F_x = 0={F_{\mathit{Bx}}}+{F_{\mathit{Ax}}}-{F_4}+{F_1}$

$\tag{2} \sum F_y = 0={F_{\mathit{By}}}+{F_{\mathit{Ay}}}-{F_3}-{F_2}$

$\tag{3} \sum M(A) = 0=6 {F_{\mathit{By}}} a+{F_4} a-4 {F_3} a-{F_2} a-2 {F_1} a$

The vertical support reaction FBy can thus be calculated directly:

$\tag{4} {F_{\mathit{By}}}=-\frac{{F_4}-4 {F_3}-{F_2}-2 {F_1}}{6}$

FBy is now substituted into equation 1, whereby FAy follows.

$\tag{5} 0={F_{\mathit{Ay}}}-\frac{{F_4}-4 {F_3}-{F_2}-2 {F_1}}{6}-{F_3}-{F_2}$

$\tag{6} {F_{\mathit{Ay}}}=\frac{{F_4}+2 {F_3}+5 {F_2}-2 {F_1}}{6}$

Next, the balance of forces in the y-direction is set up for section I in order to calculate the joint force FCy:

$\tag{7} \sum F_y = 0={F_{\mathit{Cy}}}+{F_{\mathit{Ay}}}-{F_2}$

$\tag{8} {F_{\mathit{Cy}}}=-\frac{{F_4}+2 {F_3}-{F_2}-2 {F_1}}{6}$

The sum of moments around A leads to FCx.

$\tag{9} \sum M(A) = 0=2 {F_{\mathit{Cy}}} a-3 {F_{\mathit{Cx}}} a-{F_2} a-2 {F_1} a$

$\tag{10} {F_{\mathit{Cx}}}=-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}$

The equilibrium of forces in the x-direction for section I results in the horizontal support reaction in A.:

$\tag{11} 0={F_{\mathit{Cx}}}+{F_{\mathit{Ax}}}+{F_1}$

$\tag{12} {F_{\mathit{Ax}}}=\frac{{F_4}+2 {F_3}+2 {F_2}-5 {F_1}}{9}$

FAx is now substituted into equation 2, whereby FBx follows.

$\tag{13} {F_{\mathit{Bx}}}=\frac{8 {F_4}-2 {F_3}-2 {F_2}-4 {F_1}}{9}$

The second solution leads to the desired results faster!